mbader@cac.washington.edu (Mark Bader) (08/25/89)
Does anyone know of a way to extract the "rest of the line" in awk..
e.g. I have a line that looks like
%% Heading "This is a Graph Heading"
and I want to do a {print $3} to get the "This is a Graph Heading" part,
but this obviously dosen't work. Is there a way to do this?
If there were a shift command, then I could shift off the first 2 fields,
but is there a way to simulate this?
Thanks in advance,
Mark Bader INTERNET: mbader@cac.washington.edu
Networks and Distributed Computing, UW BITNET: mbader@uwav1
3737 Brooklyn Ave. NE BELL: (206) 543-5128
Seattle, WA 98105
davidsen@crdos1.crd.ge.COM (Wm E Davidsen Jr) (08/25/89)
You can get what you want by using substring to extract from $0. -- bill davidsen (davidsen@crdos1.crd.GE.COM -or- uunet!crdgw1!crdos1!davidsen) "The world is filled with fools. They blindly follow their so-called 'reason' in the face of the church and common sense. Any fool can see that the world is flat!" - anon
karish@forel.stanford.edu (Chuck Karish) (08/25/89)
In article <3368@blake.acs.washington.edu> mbader@cac.washington.edu (Mark Bader) wrote: >I have a line that looks like >%% Heading "This is a Graph Heading" >and I want to do a {print $3} to get the "This is a Graph Heading" part, >but this obviously dosen't work. Is there a way to do this? {for (num = 3; num <= NF; num++) {printf "%s ", $num}} {printf "\n"} Chuck Karish karish@mindcraft.com (415) 493-9000 karish@forel.stanford.edu
jaap+@andrew.cmu.edu (Jaap Akkerhuis) (08/26/89)
> Excerpts from netnews.comp.unix.questions: 25-Aug-89 extracting "rest of > line" i.. Mark Bader@cac.washingto (670) > Does anyone know of a way to extract the "rest of the line" in awk.. > e.g. I have a line that looks like > %% Heading "This is a Graph Heading" > and I want to do a {print $3} to get the "This is a Graph Heading" part, > but this obviously dosen't work. Is there a way to do this? > If there were a shift command, then I could shift off the first 2 fields, but is there a way to simulate this? Allthough it is a bit ugly, you can empty the first two fields, so something like: { $1="\0"; $2="\0"; print $0 } Of course, now you will have two spaces in front of the line printed. If you want to get rid of this you need a for loop in the style of "{ for(i = 3; i < NF; i++) printf("%s ", $i); print ""}" or just filter them out with an "awk '{ $1=""; $2=""; print $0 }' | sed 's/^ //'". There are tons of other ways to do this as well. jaap
john@chinet.chi.il.us (John Mundt) (08/26/89)
In article <3368@blake.acs.washington.edu> mbader@cac.washington.edu (Mark Bader) writes: >Does anyone know of a way to extract the "rest of the line" in awk.. >e.g. I have a line that looks like > >%% Heading "This is a Graph Heading" > >and I want to do a {print $3} to get the "This is a Graph Heading" part, >but this obviously dosen't work. Is there a way to do this? If I understand this correctly, and if the quotes come around the heading all the time, you can use the split() function to break the line up in to three substriings, namely 1) <%% Heading > 2) <This is a Graph Heading> 3) <> by using split as in split($0,chop,"\""); # I don't know if an escaped # quote will work or if you'd have to # change FS to it You could then print the thing, with the quotes, by doing printf("\"%s\"\n",chop[2]); If there are no quotes, you could still split it at every space and print each separate array piece until you ran out of them: (I assume that there are a variable number of words in the heading) split($0,chop); i = 3; while (chop[i] != "") printf("%s ",chop[i++]); printf("\n"); for ( i = 0;; i++) # zero out the array at the end chop[i] = ""; # after each use The only problem here would be to allow for several spaces in a row as in "This is a heading". I haven't tested split to see if it would skip these null fields or not. -- --------------------- John Mundt Teachers' Aide, Inc. P.O. Box 1666 Highland Park, IL john@chinet.chi.il.us (312) 998-5007 (Day voice) || -432-8860 (Answer Mach) && -432-5386 Modem
steinbac@hpl-opus.HP.COM (Gunter Steinbach) (08/26/89)
How about '{$1=$2=""; print}' ?
Yes, I admit you get two extra field separators before the "rest".
Guenter Steinbach | hplabs!hpl-opus!gunter
| gunter%hpl-opus@hp-sde.hp.comdg@lakart.UUCP (David Goodenough) (08/27/89)
mbader@cac.washington.edu (Mark Bader) asks: > Does anyone know of a way to extract the "rest of the line" in awk.. > e.g. I have a line that looks like > > %% Heading "This is a Graph Heading" > > and I want to do a {print $3} to get the "This is a Graph Heading" part, > but this obviously dosen't work. Is there a way to do this? It's fairly grotesque, a lot of work, but the following should work, assuming you're not to worried about spaces. { for (i = 3; i <= NF; i++) { if (i == NF) { a = "\n"; } else { a = " "; } printf "%s%s", $i, a; } } Or somesuch. I'm making the above up as I type here, so I can't guarantee it, but it should provide a starting point. An alternative approah is to use length($1) and length($2) and substr(), but again that assumes a lot about the input record. Enough rambling, I'm not really an awk guru. We now return you to your regular news articles. -- dg@lakart.UUCP - David Goodenough +---+ IHS | +-+-+ ....... !harvard!xait!lakart!dg +-+-+ | AKA: dg%lakart.uucp@xait.xerox.com +---+
davr@hrtix.UUCP (David C. Raines) (08/28/89)
In article <9363@chinet.chi.il.us>, john@chinet.chi.il.us (John Mundt) writes: > In article <3368@blake.acs.washington.edu> mbader@cac.washington.edu (Mark Bader) writes: > >Does anyone know of a way to extract the "rest of the line" in awk.. > >e.g. I have a line that looks like > > > >%% Heading "This is a Graph Heading" > > > >and I want to do a {print $3} to get the "This is a Graph Heading" part, > >but this obviously dosen't work. Is there a way to do this? How about: for (v = 3; v <= NF; v++) printf ("%s ", $v) printf "\n" or alternatively (if the quotes are always present), set FS='"' and print $2. -- -- TCA David Raines 5 National Dr. UUCP: ...!uunet!hrtix!davr Windsor Locks, CT 06096
poage@sunny.ucdavis.edu (Tom Poage) (08/31/89)
To extract the third argument from
%% Heading "This is a Graph Heading"
try
/^%% Heading/ {
n = split($0, array, "\"")
print array[2]
}
which will produce
This is a Graph Heading
Tom.
--
Tom Poage, Clinical Engineering
Universiy of California, Davis, Medical Center, Sacramento, CA
poage@sunny.ucdavis.edu {...,ucbvax,uunet}!ucdavis!sunny!poagejon@jonlab.UUCP (Jon H. LaBadie) (09/05/89)
In article <671@lakart.UUCP>, dg@lakart.UUCP (David Goodenough) writes: > mbader@cac.washington.edu (Mark Bader) asks: > > Does anyone know of a way to extract the "rest of the line" in awk.. > > e.g. I have a line that looks like > > > > %% Heading "This is a Graph Heading" > > > > and I want to do a {print $3} to get the "This is a Graph Heading" part, > > but this obviously dosen't work. Is there a way to do this? > > It's fairly grotesque, a lot of work, but the following should work, > assuming you're not to worried about spaces. > SOLUTION DELETED New awk added a substitute function that can get rid of unwanted fields whild preserving the spacing in the rest of the record. For example, sub($1, "", $0) should get rid of field on while leaving the remainder of the record intact (including spacing). Repeating the above command should get rid of field one and two. An alternative, in one statement is sub($1 "[ ]*" $2 "[ ]*", "", $0) where each set of square brackets contains a tab and a space. Remember to preserve $1 and $2 if the logic of the program requires them at a later point in the code. Jon LaBadie -- Jon LaBadie {att, princeton, bcr}!jonlab!jon {att, attmail, bcr}!auxnj!jon
bob@wyse.wyse.com (Bob McGowen Wyse Technology Training) (09/07/89)
In article <4831@portia.Stanford.EDU> karish@forel.stanford.edu (Chuck Karish) writes: ---deleted--- >>and I want to do a {print $3} to get the "This is a Graph Heading" part, >>but this obviously dosen't work. Is there a way to do this? > >{for (num = 3; num <= NF; num++) {printf "%s ", $num}} >{printf "\n"} Or: awk -F\" '{print $2}' which sets the field separator to " and then prints the second field, which is between the two quotes. If you need the quotes output then the print could be: print "\""$2"\"" This will guarantee that everything between the quotes will be printed, while anything that may follow the second quote will be ignored. Also, no assumptions are made about the number of leading fields so modification of the data file will be less likely to require changing of the the awk script. Bob McGowan (standard disclaimer, these are my own ...) Customer Education, Wyse Technology, San Jose, CA ..!uunet!wyse!bob bob@wyse.com